Exam 6: Series Solutions of Linear Equations

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For the equation (x2−16)3(x−1)y′′−2xy′+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 , the point x=0x = 0 is

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A

For the differential equation (x2−4)2y′′−2xy′+y=0\left( x ^ { 2 } - 4 \right) ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 , the point x=0x = 0 is

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The recurrence relation for the differential equation xy′′+2y′−xy=0x y ^ { \prime \prime } + 2 y ^ { \prime } - x y = 0 is

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The differential equation x2y′′+xy′+(x2−1/16)y=0x ^ { 2 } y ^ { \prime \prime } + x y ^ { \prime } + \left( x ^ { 2 } - 1 / 16 \right) y = 0 is

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The solution of the recurrence relation in the previous problem is

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Consider the differential equation xy′′−xy′+y=0x y ^ { \prime \prime } - x y ^ { \prime } + y = 0 . The indicial equation is r(r−1)=0r ( r - 1 ) = 0 . The recurrence relation is ck+1(k+r+1)+(k+r)−ck(k+r−1)=0c _ { k + 1 } ( k + r + 1 ) + ( k + r ) - c _ { k } ( k + r - 1 ) = 0 . A series solution corresponding to the indicial root r=0r = 0 is

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The solution of the previous problem is

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The differential equation is (1−x2)y′′−2xy′+20y=0\left( 1 - x ^ { 2 } \right) y ^ { \prime \prime } - 2 x y ^ { \prime } + 20 y = 0 is

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The solution of the previous problem is

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The singular points of the differential equation xy′′+y′+y(x+2)/(x−4)=0x y ^ { \prime \prime } + y ^ { \prime } + y ( x + 2 ) / ( x - 4 ) = 0 are

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The recurrence relation for the power series solution about x=0x = 0 of the differential equation y′′+y=0y ^ { \prime \prime } + y = 0 is

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The recurrence relation for the power series solution about x=0x = 0 of the differential equation y′′−y=0y ^ { \prime \prime } - y = 0 is (for k=0,1,2,…k = 0,1,2 , \ldots )

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For the equation (x2−16)3(x−1)y′′−2xy′+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 , the point x=1x = 1 is

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The differential equation is (1−x2)y′′−2xy′+12y=0\left( 1 - x ^ { 2 } \right) y ^ { \prime \prime } - 2 x y ^ { \prime } + 12 y = 0 is

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The solution of the recurrence relation in the previous problem is

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A power series solution about x=0x = 0 of the differential equation y′′−y=0y ^ { \prime \prime } - y = 0 is

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The solution of the previous problem is

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The singular points of x2(x−1)y′′−2xy′+y=0x ^ { 2 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 are x=x = Select all that apply.

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In the previous problem, a second solution is

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For the equation (x2−16)3(x−1)y′′−2xy′+y=0\left( x ^ { 2 } - 16 \right) ^ { 3 } ( x - 1 ) y ^ { \prime \prime } - 2 x y ^ { \prime } + y = 0 , the point x=4x = 4 is

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